What is the sum from n=1 to infinity of 8(0.8)^{n-1} ?

The sum from n=1 to infinity of 8(0.8)^{n-1} is converges

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sum from n=1 to infinity of 8(0.8)^{n-1}

n = 1 \sum \infty 8 (0.8)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 8 \cdot 0. 8^{n - 1}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 8 \cdot n = 1 \sum \infty 0. 8^{n - 1}

Apply Series Ratio Test:

converges

= 8 converges

= converges

\int 105 e^{0.05 t} d t

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x \to 3 lim (x^{2} - 5)

y^{^{'}} = \frac{y + x ^{2} - y ^{2}}{x}

What is the sum from n=1 to infinity of 8(0.8)^{n-1} ?
The sum from n=1 to infinity of 8(0.8)^{n-1} is converges